Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My question is related to this one here but is different in that I am wondering about the CW structure on such a space. I am trying to put a CW structure on $S^2/S^0$ and I think that we have $1$ 0 -cell, $1$ 1 - cell and $1$ 2-cell. My $1$ - cell is just some path connecting the north pole to the south pole. However then I run into trouble because by Van - Kampen the fundamental group of $S^2/S^0$ is zero and not $\Bbb{Z}$ as stipulated in the link above.

Question: What is wrong with this cell structure on $S^2/S^0$? It seems to get $\Bbb{Z}$ I would need it to have $2$ 1 -cells but how is this obvious from the definition of $S^2/S^0$?

share|improve this question
1  
You have to attach the $1$-cell before the $2$-cell. So, if you have only one $0$-cell, there is no north (or south) pole to which to attach the $1$-cell. –  Joe Johnson 126 Nov 7 '12 at 2:16
    
Start with two $0$-cells. Then use three $1$-cells. –  Joe Johnson 126 Nov 7 '12 at 2:17
    
@JoeJohnson126 With regards to your first comment, why can't I start with one $0$ - cell? Because in the quotient space there is only one $0$ - cell no? –  user38268 Nov 7 '12 at 3:04
    
@BenjamLim: You may be able to start with one $0$-cell. But, your $2$-cells have to be attached along the $1$-skeleton. With only one $0$-cell, your $1$-skeleton will be $S^1$. –  Joe Johnson 126 Nov 7 '12 at 5:30
    
@JoeJohnson126 This is the thing right now that makes me confused because then by Van Kampen I have that $\pi_1(X) = \pi_1(S^1)/N$ where $X$ is my space now and $N$ is some normal subgroup in $\pi_1(S^1)$. I get that $N = \Bbb{Z}$ and so how do I get $\pi_1(X) = \Bbb{Z}$? –  user38268 Nov 7 '12 at 6:51
show 2 more comments

1 Answer 1

up vote 1 down vote accepted

Take the CW-decomposition to be one 0-cell and one 1-cell and one 2-cell, where the boundary of the 2-cell maps onto the 0-cell and the boundary of the 1-cell also maps onto the 0-cell. Then by van-Kampen on $X=A\cup B$, with $A$ slightly containing more than the 2-cell and $B$ slightly containing more than the 1-cell, we have $A\cap B\simeq\lbrace\text{0-cell}\rbrace$ and $\pi_1A\cong\pi_1S^2=0$ and $\pi_1B\cong\pi_1S^1=\mathbb{Z}$ and hence $\pi_1X\cong\mathbb{Z}$.

The problem with your given cell structure is that you haven't actually told us the attaching maps... for instance, where is your 1-cell attaching to? By definition, it must attach to cells of dimension $<1$, i.e. to your given 0-cell.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.